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Chapter 15: Problem 9

Find a parametrization of the surface. (There are many correct ways to do these, so your answers may not be the same as those in the back of the book.) The surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=2,\) and \(z=0\)

Short Answer

Expert verified
The surface is parameterized by \( \mathbf{r}(u, v) = \langle u, v, 4 - v^2 \rangle \) with \( 0 \leq u \leq 2 \) and \( -2 \leq v \leq 2 \).

Step by step solution

01

Understand the Surface and Planes

The surface given is a parabolic cylinder described by the equation \( z = 4 - y^2 \). It is cut by the planes \( x = 0 \), \( x = 2 \), and \( z = 0 \). This implies we are considering part of the cylinder between \( x = 0 \) and \( x = 2 \), bounded below by the plane \( z = 0 \).
02

Set up the Parameters

Choose parameters \( x \) and \( y \) to describe points on the surface. Since \( x \) is bound between two planes, we can set \( x = u \) where \( 0 \leq u \leq 2 \). \( y \) is a free variable of the surface, so let \( y = v \).
03

Express z in terms of Parameters

Using the surface equation \( z = 4 - y^2 \) and our parameter \( y = v \), express \( z \) as a function of the parameter \( v \): \( z = 4 - v^2 \). Also, since the surface is cut by \( z = 0 \), \( v \) should satisfy \( 4 - v^2 \geq 0 \). Thus, the range for \( v \) is \( -2 \leq v \leq 2 \) because the maximum value of \( |v| \) occurs when \( z = 0 \).
04

Write the Parametrization

With our parameters identified, the parametrization of the surface can be written as a vector function: \(\mathbf{r}(u, v) = \langle u, v, 4 - v^2 \rangle \), where \( 0 \leq u \leq 2 \) and \( -2 \leq v \leq 2 \). This ensures all points on the cut surface are represented.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabolic Cylinder
A parabolic cylinder is a fascinating type of surface. It is formed by taking a parabola, which is a U-shaped curve, and extending it infinitely in one direction. In simpler terms, imagine a piece of paper folded into a parabola that you stretch into a tunnel!
A standard parabolic cylinder can be described by an equation like \( z = ax^2 + by^2 + c \). In the exercise, the equation given is \( z = 4 - y^2 \). This tells us that the parabola is parallel to the x-axis while opening downward, creating a curve up and down as y changes.
More specifically:
  • The term \( 4 - y^2 \) indicates it opens downwards because of the negative \( y^2 \) term.
  • Since \( x \) is not in the equation for \( z \), it implies that the surface is infinitely extending along the x-direction.
Understanding this simple structure helps in visualizing how the parabolic cylinder interacts with other geometrical entities like planes.
Parametrization
Parametrization is a brilliant technique used in mathematics to describe a surface using a pair of variables. It turns complex equations into simpler vector expressions.
In this context, to parametrize the parabolic cylinder, we choose two parameters to identify each point on the surface clearly. The common parameters are \( u \) and \( v \):
  • Let \( x = u \), because \( x \) operates between two defined limits or planes: \( x = 0 \) and \( x = 2 \).
  • Let \( y = v \), as it is the free variable in the surface equation, representing the vertical changes.
  • Finally, substitute \( y = v \) into the original surface equation \( z = 4 - y^2 \), giving us \( z = 4 - v^2 \).
Through these parameters, we not only simplify but also easily define the scale and extent of the surface segment involved by specifying \( 0 \leq u \leq 2 \) and \( -2 \leq v \leq 2 \). This method ensures clarity and precision in mapping out the given geometric section.
Vector Function
A vector function is a handy way to express geometric figures in mathematics. It takes parameters like numbered positions and creates a vector that represents points on a surface.
In the exercise, once we know the parameters \( u \) and \( v \), we form a vector function to fully describe the surface of the parabolic cylinder intersected by the planes:
  • Using the parameters, our vector function becomes \( \mathbf{r}(u, v) = \langle u, v, 4 - v^2 \rangle \).
  • This notation means at any point defined by \( u \) and \( v \), the vector gives the position on the surface in terms of \( x \), \( y \), and \( z \).
  • It combines all the steps in parametrization, representing the real 3D space simplistically.
This vector function format is particularly advantageous since it provides a concise and effective way to tackle multidimensional problems. It is a bridge that links algebraic concerns with spatial visualization, essential for understanding the geometry of surfaces like the parabolic cylinder.

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Most popular questions from this chapter

The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) of the flow's velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region \(R\) (no holes or missing points) and that if \(M_{x}+N_{y} \neq 0\) throughout \(R,\) then none of the streamlines in \(R\) is closed. In other words, no particle of fluid ever has a closed trajectory in \(R .\) The criterion \(M_{x}+N_{y} \neq 0\) is called Bendixson's criterion for the nonexistence of closed trajectories.

Find the flux of the field \(\mathbf{F}(x, y, z)=4 x \mathbf{i}+4 y \mathbf{j}+2 \mathbf{k}\) outward (away from the \(z\) -axis) through the surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=1\).

Use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve C. Perform the following CAS steps. a. Plot \(C\) in the \(x y\) -plane. b. Determine the integrand \((\partial N / \partial x)-(\partial M / \partial y)\) for the tangential form of Green's Theorem. c. Determine the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. \(\mathbf{F}=x^{-1} e^{y} \mathbf{i}+\left(e^{y} \ln x+2 x\right) \mathbf{j}\) C: The boundary of the region defined by \(y=1+x^{4}\) (below) and \(y=2(\text { above })\)

Work and area Suppose that \(f(t)\) is differentiable and positive for \(a \leq t \leq b .\) Let \(C\) be the path \(\mathbf{r}(t)=t \mathbf{i}+f(t) \mathbf{j}, a \leq t \leq b\) and \(\mathbf{F}=\) yi. Is there any relation between the value of the work integral $$ \int_{C} \mathbf{F} \cdot d \mathbf{r} $$ and the area of the region bounded by the \(t\) -axis, the graph of \(f\) and the lines \(t=a\) and \(t=b ?\) Give reasons for your answer.

The tangent plane at a point \(P_{0}\left(f\left(u_{0}, v_{0}\right), g\left(u_{0}, v_{0}\right), h\left(u_{0}, v_{0}\right)\right)\) on a parametrized surface \(\mathbf{r}(u, v)=f(u, v) \mathbf{i}+g(u, v) \mathbf{j}+h(u, v) \mathbf{k}\) is the plane through \(P_{0}\) normal to the vector \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right) \times \mathbf{r}_{v}\left(u_{0}, v_{0}\right),\) the cross product of the tangent vectors \(\mathbf{r}_{u}\left(u_{0}, v_{0}\right)\) and \(\mathbf{r}_{2}\left(u_{0}, v_{0}\right)\) at \(P_{0} .Find an equation for the plane tangent to the surface at \)P_{0} .\( Then find a Cartesian equation for the surface and sketch the surface and tangent plane together. The cone \)\mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+r \mathbf{k}, r \geq 0\( \)0 \leq \theta \leq 2 \pi\( at the point \)P_{0}(\sqrt{2}, \sqrt{2}, 2)\( corresponding to \)(r, \theta)=(2, \pi / 4)$
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